{"id":5565,"date":"2024-01-07T02:03:15","date_gmt":"2024-01-06T20:33:15","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=5565"},"modified":"2024-01-10T16:00:40","modified_gmt":"2024-01-10T10:30:40","slug":"ts-inter-1st-year-maths-1b-solutions-chapter-6-ex-6b","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1b-solutions-chapter-6-ex-6b\/","title":{"rendered":"TS Inter 1st Year Maths 1B Solutions Chapter 6 Direction Cosines and Direction Ratios Ex 6(b)"},"content":{"rendered":"

Students must practice these TS Intermediate Maths 1B Solutions<\/a> Chapter 6 Direction Cosines and Direction Ratios Ex 6(b) to find a better approach to solving the problems.<\/p>\n

TS Inter 1st Year Maths 1B Direction Cosines and Direction Ratios 6(b)<\/h2>\n

Question 1.
\nFind the direction ratios of the line joining the points (3, 4, 0) and (4, 4, 4). (V.S.A.Q.)
\nAnswer:
\nLet A = (3, 4, 0) and B = (4, 4, 4) be the given points.
\nThen d.r.’s of AB = ( 4 – 3, 4 – 4, 4 – 0) = (1, 0, 4)<\/p>\n

Question 2.
\nThe direction ratios of a line are (-6, 2, 3). Find its direction cosines. (V.S.A.Q.)
\nAnswer:
\nd.r’s of the line are -6, 2, 3.
\n\u2234 \\(\\sqrt{a^2+b^2+c^2}\\) = \\(\\sqrt{36+4+9}\\) = \u221a49 = 7
\nd.c.’s of the line are \\(\\frac{-6}{7}, \\frac{2}{7}, \\frac{3}{7}\\).<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the cosine of the angle between the lines whose direction cosines are
\n\\(\\left(\\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}\\right)\\) and \\(\\left(\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}, 0\\right)\\).
\nAnswer:
\ncos \u03b8 = l1<\/sub> l2<\/sub> + m1<\/sub> m2<\/sub> + n1<\/sub>n2<\/sub>
\n\"TS<\/p>\n

Question 4.
\nFind the angle between the lines whose d.r.’s are (1, 1, 2), (\u221a3 , – \u221a3, 0). (V.S.A.Q.)
\nAnswer:
\nWe have
\n\"TS<\/p>\n

Question 5.
\nShow that the lines with direction cosines \\(\\left(\\frac{12}{13}, \\frac{-3}{13}, \\frac{-4}{13}\\right)\\) and \\(\\left(\\frac{4}{13}, \\frac{12}{13}, \\frac{3}{13}\\right)\\) are Perpendicular to each other. (V.S.A.Q.)
\nAnswer:
\nWe have the condition for two lines with d.c.’s (l1<\/sub>, m1<\/sub>, n1<\/sub>) and (l2<\/sub>, m2<\/sub>, n2<\/sub>) to be perpendicular is l1<\/sub>l2<\/sub> + m1<\/sub>m2<\/sub> + n1<\/sub>n2<\/sub> = 0
\n\u2234 l1<\/sub>l2<\/sub> + m1<\/sub>m2<\/sub> + n1<\/sub>n2<\/sub>
\n\"TS
\n\u2234 The given lines are perpendicular.<\/p>\n

Question 6.
\nO is the origin, P(2, 3, 4) and Q (1, k, 1) are points such that OP \u22a5 OQ . Find k.
\nAnswer:
\nd.r.’s of OP = 2, 3, 4
\nd.r.’s of OQ = 1, k, 1
\nOP and OQ are perpendicular.
\n\u21d2 a1<\/sub>a2<\/sub> + b1<\/sub>b2<\/sub> + c1<\/sub>c2<\/sub> = 0
\n\u21d2 2 + 3k + 4 = 0 \u21d2 3k + 6 = 0 \u21d2 k = – 2<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\nIf the direction ratios of a line are (3, 4, 0) find its direction cosines and also the angles made with the coordinate axes. (S.A.Q.)
\nAnswer:
\nd.c.\u2019s of the line are (3, 4, 0).
\n\\(\\sqrt{a^2+b^2+c^2}\\) = \\(\\sqrt{9+16}\\) = 5
\n\u2234 d.c.’s of the line are
\nIf \u03b1, \u03b2, \u03b3 are angles made by the line with the
\ncoordinate axes then cos \u03b1 = \\(\\frac{3}{5}\\), cos \u03b2 = \\(\\frac{4}{5}\\), cos \u03b3 = 0
\n\u2234 \u03b1 = cos-1<\/sup> (3\/5), \u03b2 = cos-1<\/sup> (4\/5), \u03b3 = \\(\\frac{\\pi}{2}\\)
\n\u2234 Angles made with coordinate axes are
\ncos-1<\/sup> (3\/5), cos-1<\/sup> (4\/5) and \\(\\frac{\\pi}{2}\\).<\/p>\n

Question 2.
\nShow that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6). (S.A.Q.)
\nAnswer:
\nLet A = (1, -1, 2), B = (3, 4, -2), C = (0, 3, 2) and D = (3, 5, 6) be the given points.
\nd.r.’s of AB = (3 – 1, 4 + 1, -2 – 2) = (2, 5,4)
\nd.r.’s of CD = (3 – 0, 5 – 3, 6 – 2) = (3, 2, 4)
\n\u2234 a1<\/sub> a2<\/sub> + b1<\/sub> b2<\/sub> + c1<\/sub> c2<\/sub> = 2(3) + 5(2) + (-4) (4) = 0
\n\u2234 AB and CD are perpendicular.<\/p>\n

Question 3.
\nFind the angle between DC and AB where A = (3, 4, 5), B = (4, 6, 3), C = (-1, 2, 4) and D = (1, 0, 5). (S.A.Q.)
\nAnswer:
\nd.r.’s of AB are (4 – 3, 6 – 4, 3 – 5) = (1, 2, -2)
\nd.r.’s of CD are (1 + 1, 0 – 2, 5 – 4) = (2, -2, 1)
\n\"TS<\/p>\n

Question 4.
\nFind the direction cosines of a line which is perpendicular to the lines, whose direction ratios are (1, -1, 2) and (2, 1, -1). (S.A.Q.)
\nAnswer:
\nLet the d.c.’s of the required line be a, b, c. This is perpendicular to the line whose d.r.’s are (1,-1, 2) and (2, 1, -1).
\nThen a – b + 2c = 0
\nand 2a + b – c = 0
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 5.
\nShow that the points (2, 3, -4), (1, -2 ,3) and (3, 8, -11) are collinear. (S.A.Q.)
\nAnswer:
\nLet A = (2, 3, -4), B = (1, -2, 3) and C = (3, 8, -11) be the three given points.
\n\"TS
\n\u2235 AB + AC = 5\u221a3 + 5\u221a3 = 10\u221a3 = BC
\nWe have A, B, C are collinear.<\/p>\n

Question 6.
\nShow that the points (4, 7, 8), (2, 3, 4), (-1, -2, 1), (1, 2, 5) are the vertices of a parallelogram. (S.A.Q.)
\nAnswer:
\nLet A = (4, 7, 8), B = (2, 3, 4), C = (-1, -2, 1) and D = (1, 2, 5) be the four given points.
\n\"TS
\nd.r.’s of \\(\\overline{\\mathrm{AB}}\\) = (2 – 4, 3 – 7, 4 – 8)
\n= (-2, -4, -4) ………………. (1)
\nd.r.’s of \\(\\overline{\\mathrm{DC}}\\) = (- 1 – 1, – 2 – 2, 1 – 5)
\n= (-2, -4, -4) ………………… (2)
\nd.r.\u2019s of \\(\\overline{\\mathrm{AB}}\\) = DR’s of \\(\\overline{\\mathrm{DC}}\\) we have \\(\\overline{\\mathrm{AB}}\\) is parallel to \\(\\overline{\\mathrm{DC}}\\).
\nd.r.’s of \\(\\overline{\\mathrm{AD}}\\) = (1 – 4, 2 – 7, 5 – 8)
\n= (-3, -5, -3) ……………………. (3)
\nd.r.’s of \\(\\overline{\\mathrm{BC}}\\) = (-1 – 2, -2 – 3, 1 – 4)
\n= (-3, -5, -3) …………………….. (4)
\n\u2235 d.r.’s of \\(\\overline{\\mathrm{AD}}\\) = d.r.\u2019s of \\(\\overline{\\mathrm{BC}}\\), we have \\(\\overline{\\mathrm{AD}}\\) is parallel to \\(\\overline{\\mathrm{BC}}\\).
\nFrom (1) and (3),
\n(-2) (-3) + (-4) (-5) + (-4) (-3) \u2260 0
\nFrom (2) and (4),
\n(-2) (-3) + (-4) (-5) + (-4) (-3) \u2260 0
\nWe have \\(\\overline{\\mathrm{AD}}\\) is not perpendicular to \\(\\overline{\\mathrm{AB}}\\) and \\(\\overline{\\mathrm{DC}}\\) is not perpendicular to \\(\\overline{\\mathrm{BC}}\\).
\nAlso d.r.’s of \\(\\overline{\\mathrm{AC}}\\) = (-1 – 4, -2 -7, 1 – 8)
\n= (-5, -9, -7) ………………….. (5)
\nd.r.’s of B\\(\\overline{\\mathrm{BD}}\\) = (1 – 2, 2 – 3, 5 – 4)
\n= (-1, -1, 1) ………………….. (6)
\nFrom (5) and (6)
\n(- 5) (- 1) + (- 9) (- 1) + (1) (-7) \u2260 0
\nHence diagonals \\(\\overline{\\mathrm{AC}}\\) and \\(\\overline{\\mathrm{BD}}\\) are not perpendicular. Hence ABCD is a parallelogram.<\/p>\n

III.
\nQuestion 1.
\nShow that the lines whose direction cosines are given by l + m + n = 0, 2mn + 3nl – 5lm= 0 are perpendicular to each other.(E.Q.) (March \u201912)
\nAnswer:
\nGiven l + m + n = 0 ………………… (1)
\nand 2mn + 3nl – 5lm = 0 …………………….. (2)
\nFrom (1), l = – (m + n)
\n\u2234 From (2), 2mn – 3n(m + n) + 5m(m + n) = 0
\n\u21d2 2mn – 3mn – 3n2<\/sup> + 5m2<\/sup> + 5mn = 0
\n\u21d2 5m2<\/sup> + 4mn – 3n2<\/sup> = 0
\n\u21d2 5\\(\\left(\\frac{m}{n}\\right)^2\\) + 4\\(\\left(\\frac{m}{n}\\right)\\) – 3 = 0
\nThis is a quadratic equation in \\(\\left(\\frac{m}{n}\\right)\\) and let \\(\\frac{m_1}{n_1}\\) and \\(\\frac{m_2}{n_2}\\) be the roots of the quadratic equation.
\nThen product of roots \\(\\frac{m_1}{n_1} \\cdot \\frac{m_2}{n_2}=\\frac{-3}{5}\\)
\n\u21d2 \\(\\frac{m_1 m_2}{3}=\\frac{n_1 n_2}{-5}\\) = k (suppose) …………… (3)
\nFrom (1), m = – (l + n)
\n– 2n (l + n) + 3nl + 5l (l + n) = 0
\n\u21d2 5l2<\/sup> + 6nl – 2n2<\/sup> = 0
\n\"TS
\n\u2234 l1<\/sub>l2<\/sub> = 2k, m1<\/sub>m2<\/sub> = 3k, n1<\/sub>n2<\/sub> = – 5k
\n\u2234 l1<\/sub>l2<\/sub> + m1<\/sub>m2<\/sub> + n1<\/sub>n2<\/sub> = 0
\n\u2234 The two lines are perpendicular.<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l2<\/sup> + m2<\/sup> – n2<\/sup> = 0. (E.Q.) (May 2014, ‘11,2007, Mar. ’13, ’07, June 2004)
\nAnswer:
\nGiven equations are
\nl + m + n = 0 ………………….. (1) and
\nl2<\/sup> + m2<\/sup> – n2<\/sup> = 0 …………………… (2)
\nFrom (1), l = – (m + n)
\nFrom (2), [-(m + n)]2<\/sup> + m2<\/sup> – n2<\/sup> = 0
\n\u21d2 m2<\/sup> + n2<\/sup> + 2mn + m2<\/sup> – n2<\/sup> = 0
\n\u21d2 2m2<\/sup> + 2mn = 0
\n\u21d2 2m (m + n) = 0
\n\u21d2 m = 0 or m = -n<\/p>\n

Case (i): If m = 0 then
\nl(0) + m (1) + n(0) = 0
\nand l + m + n = 0 ……………… (1)
\nSolving
\n\"TS
\n\u2234 d.c.’s of one pair of lines
\n(l1<\/sub>, m1<\/sub>, n1<\/sub>) = \\(\\left(\\frac{1}{\\sqrt{2}}, \\frac{0}{\\sqrt{2}}, \\frac{-1}{\\sqrt{2}}\\right)\\)<\/p>\n

Case (ii): If m + n = 0 then
\nl(0) + m(1) + n(1) = 0
\nand l + m + n = 0
\nSolving
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nIf a ray makes angles \u03b1, \u03b2, \u03b3 and \u03b4 with the four diagonals of a cube find a cube find cos2<\/sup> \u03b1 + cos2<\/sup> \u03b2 + cos2<\/sup> \u03b3 + cos2<\/sup> \u03b4 (E.Q.) (March 2005, May 2005)
\nAnswer:
\n\"TS
\nLet the side of the cube be of length ‘a’. Let one of the vertices of the cube through the origin ‘O’ and axes be along the three edges \\(\\overline{\\mathrm{OA}}\\), \\(\\overline{\\mathrm{OB}}\\) and \\(\\overline{\\mathrm{OC}}\\). The diagonals of the cube are OP, CD, AE and BF d.r.’s of the diagonals are (a, a, a), (a, a, -a), (-a, a, a) and (a, -a, a) respectively.
\nLet the d.c.’s of the given ray be l, m, n. If this make angles \u03b1, \u03b2, \u03b3 and \u03b4 with the four diagonals of the cube then
\n\"TS<\/p>\n

Question 4.
\nIf (l1<\/sub>, two intersecting lines. Show that the d.c’s of two lines bisecting the angles between them are proportioned to l1<\/sub> \u00b1 l2<\/sub>, m1<\/sub> \u00b1 m2<\/sub>, n1<\/sub> \u00b1 n2<\/sub> (E.Q.)
\nAnswer:
\nLet OA, OB be the given lines and A, B be the points at which distances from 0.
\n\u21d2 Coordinates of A = (l1<\/sub>, m1<\/sub>, n1<\/sub>) and B = (l2<\/sub>, m2<\/sub>, n2<\/sub>)
\nMid point of AB, is:
\nP = \\(\\left(\\frac{l_1+l_2}{2}, \\frac{m_1+m_2}{2}, \\frac{n_1+n_2}{2}\\right)\\)
\n\u2234 OP is the bisector of \u2220AOB \u21d2 d.r,’s of OP are l1<\/sub> + l2<\/sub>, m1<\/sub> + m2<\/sub>, n1<\/sub> + n2<\/sub>
\nNow coordinates of B = (-l2<\/sub>, -m2<\/sub>, -n2<\/sub>) and mid point of AB’ is
\nQ = \\(\\left(\\frac{l_1-l_2}{2}, \\frac{m_1-m_2}{2}, \\frac{n_1-n_2}{2}\\right)\\)
\n\u2234 OQ is a bisector of \u2220AOB \u21d2 d.r.’s of OQ are l1<\/sub> – l2<\/sub>, m1<\/sub> – m2<\/sub>, n1<\/sub> – n2<\/sub><\/p>\n

Question 5.
\nA (-1, 2, -3), B (5, 0, -6), C (0, 4, -1) are three points. Show that the direction cosines of the bisector of \u2220BAC are proportional to (25, 8, 5) and (-11, 20, 23). (E.Q.)
\nAnswer:
\nGiven A (-1, 2, -3), B (5, 0, -6) and C (0,4, -1) are three points.
\nd.r.’s of AB are = (5 + 1, 0 – 2, – 6 + 3) = (6, – 2, – 3)
\n\"TS
\nHence d.r.’s of other bisector are (-11, 20, 23) Hence direction cosines of the bisectors of \u2220BAC are proportional to (25, 8, 5) and (-11, 20, 23).<\/p>\n

\"TS<\/p>\n

Question 6.
\nIf (6, 10, 10), (1, 0, -5), (6, – 10, 0) are vertices of a triangle. Find the direction ratios of its sides. Determine whether it is fight angled or isoceles. (S.A.)
\nAnswer:
\nLet A (6,10,10), B (1, 0, -5) and C (6, -10, 0) are the vertices of \u2206ABC
\nd.r.’s of AB = (-5, -10, -15) \u21d2 (1,2, 3)
\nd.r.’s of BC = (5, -10, -5) \u21d2 (1, -2, 1)
\nd.r.’s of AC = (0, 20, 10) \u21d2 (0, 2, 1)
\n\"TS
\n\u2234 The given triangle is a right angled triangle.<\/p>\n

Question 7.
\nThe vertices of a triangle are A (1, 4, 2), B (-2, 1, 2), C (2, 3, -4). Find \u2220A, \u2220B, \u2220C. (S.A.Q.)
\nAnswer:
\nGiven A (1, 4, 2), B (-2, 1, 2) and C (2, 3, -4) are the vertices of ABC.
\nd.r.’s of AB are = 3, 3, 0 i.e., 1, 1, 0
\nd.r.’s of BC are = -4, -2, 6 i.e., 2, 1,- 3
\nd.r.’s of AC are – 1, 1, 6
\n\"TS<\/p>\n

Question 8.
\nFind the angle between the lines whose direction cosines are given by the equation 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0 (E.Q.) ( May’06, ’12))
\nAnswer:
\nGiven 3l + m + 5n = 0 ……………………. (1)
\nand 6mn – 2nl + 5lm = 0 ……………………. (2)
\nFrom (1), m = – (3l + 5n)
\nFrom (2)
\n– 6n (31 + 5n) – 2n\/ – 5\/ (31 + 5n) = 0
\n\u21d2 – 18nl – 30n2<\/sup> – 2nl – 15l2<\/sup> – 25ln = 0
\n\u21d2 – 15l2<\/sup> – 45ln – 30n2<\/sup> = 0
\n\u21d2 l2<\/sup> + 3ln + 2n2<\/sup> = 0
\n\u21d2 (l + 2n) (l + n) = 0
\n\u21d2 l + 2n = 0 or l + n = 0
\n\u21d2 l = – n or l = – 2n
\nIf l = -n then m = 3n – 5n = – 2n
\n\u21d2 l : m : n = -n : -2n : n
\n= 1 : 2 : – 1
\nIf l = -2n then m = 6n – 5n = n
\n\u21d2 l : m : n = -2n : n : n = 2 : -1 : -1
\nHence d.r.’s of two lines are (1, 2, -1) and (2, -1,-1).
\nIf \u03b8 is the angle between the two lines then
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 9.
\nIf a variable line in two adjacent positions has direction cosines (l, m, n) and (l + \u03b4l, m + \u03b4m, n + \u03b4n), show that the small angle \u03b4\u03b8 between two positions is given by (\u03b4\u03b8)2<\/sup> = (\u03b4l)2<\/sup> + (\u03b4m)2<\/sup> + (\u03b4n)2<\/sup>. (E.Q.)
\nAnswer:
\nGiven direction cosines of a variable line in two adjacent positions are (l, m, n) and (l + \u03b4l, m + \u03b4m, n + \u03b4n)
\nWe have l2<\/sup> + m2<\/sup> + n2<\/sup> = 1 …………………… (1)
\nand (l + \u03b4l)2<\/sup> + (m + \u03b4m)2<\/sup> + (n + \u03b4n)2<\/sup> = 1 ………………… (2)
\nFrom (2) – (1) we have
\n(l + \u03b4l)2<\/sup> + (m + \u03b4m)2<\/sup> +(n + \u03b4n)2<\/sup> – l2<\/sup> – m2<\/sup> – n2<\/sup> = 0
\n\u21d2 2 (l. \u03b4l + m . \u03b4m + n . \u03b4n) = – [(\u03b4l)2<\/sup> + (\u03b4m)2<\/sup> +(\u03b4n)2<\/sup>]
\n\u21d2 (\u03b4l)2<\/sup> + (\u03b4m)2<\/sup> + (\u03b4n)2<\/sup> = – 2 (l\u03b4l + m\u03b4m + n\u03b4n) …………………… (3)
\nNow angle between two adjacent sides
\ncos \u03b4\u03b8 = l (l + \u03b4l) + m (m + \u03b4m) + n (n + \u03b4n)
\n= (l2<\/sup> + m2<\/sup> + n2<\/sup>) + l.\u03b4l + m.\u03b4m + n.\u03b4n
\n= 1 + l.\u03b4l + m.\u03b4m + n.\u03b4n
\n= 1 – \\(\\frac{1}{2}\\) [(\u03b4l)2<\/sup> + (\u03b4m)2<\/sup> +(\u03b4n)2<\/sup>]
\n\u2234 (\u03b4l)2<\/sup> + (\u03b4m)2<\/sup> + (\u03b4n)2<\/sup> = (1 – cos \u03b4\u03b8)
\nSince \u03b4\u03b8 is very small, sin\\(\\left(\\frac{\\delta \\theta}{2}\\right)\\) = \\(\\frac{\\delta \\theta}{2}\\)
\n\u2234 (\u03b4l)2<\/sup> + (\u03b4m)2<\/sup> + (\u03b4n)2<\/sup> = 4\\(\\frac{(\\delta \\theta)^2}{4}\\) = (\u03b4\u03b8)2<\/sup>
\n[\u2235 1 – cos\u03b8 = 2 sin2<\/sup>\\(\\frac{\\theta}{2}\\)]
\n\u2234 (\u03b4\u03b8)2<\/sup> = (\u03b4l)2<\/sup> + (\u03b4m)2<\/sup> + (\u03b4n)2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these TS Intermediate Maths 1B Solutions Chapter 6 Direction Cosines and Direction Ratios Ex 6(b) to find a better approach to solving the problems. TS Inter 1st Year Maths 1B Direction Cosines and Direction Ratios 6(b) Question 1. Find the direction ratios of the line joining the points (3, 4, 0) and … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5565"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=5565"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5565\/revisions"}],"predecessor-version":[{"id":5996,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/5565\/revisions\/5996"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=5565"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=5565"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=5565"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}